Ta có: \(\left(x^2+3x+1\right)\left(x^2+3x-3\right)-5\)

\(=\left(x^2+3x\right)^2-3\left(x^2+3x\right)+\left(x^2+3x\right)-3-5\)

\(=\left(x^2+3x\right)^2-2\left(x^2+3x\right)-8\)

\(=\left(x^2+3x-4\right)\left(x^2+3x+2\right)=\left(x+4\right)\left(x-1\right)\left(x+1\right)\left(x+2\right)\)

\(S=1^3+2^3+3^3+...+n^3=\left(1+2+3+...+n\right)^2\)

\(=\left[\dfrac{n\left(n+1\right)}{2}\right]^2=\dfrac{n^2\cdot\left(n+1\right)^2}{4}\)

Hằng đẳng thức đó là \(\left(a-b\right)^2=\left(b-a\right)^2\)

thằng nghiện game

x^5-1

=(x-1)(x^4+x^3+x^2+x+1).

-(a - 3)² = -(a² - 6a + 9) = -a² + 6a - 9

(x - 2)(x + 2) = x² - 4

-(5 + 4y)(5 - 4y) = -(25 - 16y²) = -25 + 16y²

(\(\dfrac{1}{2}\)x + 2y)(\(\dfrac{1}{2}\)x - 2y) = \(\dfrac{1}{4}\)x² - 4y²

a) \(A=\left(x^2-10x+25\right)\)\(-28\)

   \(A=\left(x-5\right)^2-28\)\(>=\)-28

MinA = -28 <=> x-5=0 <=> x=5

b)\(B=-\left(x^2+2x+1\right)+6\)

   \(B=-\left(x+1\right)^2+6\)\(< =\)6

MaxB = 6 <=> x+1=0 <=> x=-1

c)\(C=-5\left(x^2-\frac{6}{5}x+\frac{9}{25}\right)-\frac{26}{5}\)

   \(C=-5\left(x-\frac{3}{5}\right)^2-\frac{26}{5}\)\(< =-\frac{26}{5}\)

MaxC = \(-\frac{26}{5}\)<=> \(x-\frac{3}{5}=0\)<=> x=\(\frac{3}{5}\)

d)\(D=-3\left(x^2+\frac{1}{3}x+\frac{1}{36}\right)+\frac{61}{12}\)

\(D=-3\left(x+\frac{1}{6}\right)^2+\frac{61}{12}\)\(< =\frac{61}{12}\)

MacD = \(\frac{61}{12}\)<=> \(x+\frac{1}{6}=0\)<=> \(x=\frac{-1}{6}\)

Đúng thì nhớ tích cho minh nha

\(x^3-8+12x-6x^2\)

\(=x^3-3.x^2.2+3.x.2^2-2^3\)

\(=\left(x-2\right)^3\)